10.1 Concepts of WDM
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be specified to have a wavelength of 1557.363 ± 0.005 nm (or equivalently, 192.50
THz).
Example 10.2 Consider a spectral band of 0.8 nm (or equivalently, a mean frequency
spacing of 100 GHz at a 1550-nm wavelength) within which lasers with narrow
linewidths are transmitting. How many of such signal channels fit into (a) the C
band, and (b) the combined S-band and C-band?
Solution: (a) Because the C-band ranges from 1530 to 1565 nm, one can have N =
(35 nm)/(0.8 nm per channel) = 43 independent signal channels.
(b) Because the combined S-band and C-band cover the 1460- to-1565 nm range,
one can have N = (105 nm)/(0.8 nm per channel) = 131 independent signal channels.
Example 10.3 Assume that a 16-channel WDM system has a uniform channel
spacing ν = 200 GHz and let the frequency ν n correspond to the wavelength
λ n . Let the wavelength λ 1 = 1550 nm. Calculate the wavelength spacing between
the first two channels (channels 1 and 2) and between the last two channels (channels
15 and 16). From the result, what can be concluded about using an equal-wavelength
spacing definition in this wavelength band instead of the standard equal-frequency
channel spacing specification?
Solution: To find the wavelength spacing between the first two channels (channels 1
and 2), first look at the frequency difference between these channels. The frequency
for channel N is given by ν N = ν 1 + (N − 1)ν, where ν = 200 GHz. Therefore,
ν 1 = c/λ 1 = (3 × 10
8 m/s)/(1550 nm) = 193.5483 THz yields.
λ 2 = c/ν 2 = c/[ν 1 + (2 − 1)ν] = (3 × 10
8 m/s)/[193.5483 + 0.2 THz] =
1548.40 nm. Thus between λ 1 and λ 2 the channel spacing is λ = 1.60 nm. Similarly.
λ 15 = c/ν 15 = c/[ν 1 + (15 − 1)ν] = (3 × 10
8 m/s)/[193.5483 + 2.8] =
1527.90 nm. and
λ 16 = c/ν 16 = c/[ν 1 + (16 − 1)ν] = (3 × 10
8 m/s)/[193.5483 + 3.0] =
1526.34 nm. Thus between λ 15 and λ 16 the channel spacing is λ = 1.55 nm.
This shows that the spacing between different adjacent wavelengths is not uniform.
Because the channel spacing difference at the wavelength extremes in this case is
small (about 3%), an equal-wavelength approximation is close to an equal-frequency
spacing definition.
10.1.2 Standards for WDM
Because WDM is essentially frequency division multiplexing at optical carrier
frequencies, the WDM standards developed by the International Telecommunication
Union (ITU) specify channel spacing in terms of frequency [7–10]. A key reason for
selecting a frequency spacing that is fixed, rather than a wavelength spacing that is
constant, is that when locking a laser to a particular operating mode, it is the frequency
of the laser that is fixed. Recommendation G.692 was the first ITU-T specification
387
be specified to have a wavelength of 1557.363 ± 0.005 nm (or equivalently, 192.50
THz).
Example 10.2 Consider a spectral band of 0.8 nm (or equivalently, a mean frequency
spacing of 100 GHz at a 1550-nm wavelength) within which lasers with narrow
linewidths are transmitting. How many of such signal channels fit into (a) the C
band, and (b) the combined S-band and C-band?
Solution: (a) Because the C-band ranges from 1530 to 1565 nm, one can have N =
(35 nm)/(0.8 nm per channel) = 43 independent signal channels.
(b) Because the combined S-band and C-band cover the 1460- to-1565 nm range,
one can have N = (105 nm)/(0.8 nm per channel) = 131 independent signal channels.
Example 10.3 Assume that a 16-channel WDM system has a uniform channel
spacing ν = 200 GHz and let the frequency ν n correspond to the wavelength
λ n . Let the wavelength λ 1 = 1550 nm. Calculate the wavelength spacing between
the first two channels (channels 1 and 2) and between the last two channels (channels
15 and 16). From the result, what can be concluded about using an equal-wavelength
spacing definition in this wavelength band instead of the standard equal-frequency
channel spacing specification?
Solution: To find the wavelength spacing between the first two channels (channels 1
and 2), first look at the frequency difference between these channels. The frequency
for channel N is given by ν N = ν 1 + (N − 1)ν, where ν = 200 GHz. Therefore,
ν 1 = c/λ 1 = (3 × 10
8 m/s)/(1550 nm) = 193.5483 THz yields.
λ 2 = c/ν 2 = c/[ν 1 + (2 − 1)ν] = (3 × 10
8 m/s)/[193.5483 + 0.2 THz] =
1548.40 nm. Thus between λ 1 and λ 2 the channel spacing is λ = 1.60 nm. Similarly.
λ 15 = c/ν 15 = c/[ν 1 + (15 − 1)ν] = (3 × 10
8 m/s)/[193.5483 + 2.8] =
1527.90 nm. and
λ 16 = c/ν 16 = c/[ν 1 + (16 − 1)ν] = (3 × 10
8 m/s)/[193.5483 + 3.0] =
1526.34 nm. Thus between λ 15 and λ 16 the channel spacing is λ = 1.55 nm.
This shows that the spacing between different adjacent wavelengths is not uniform.
Because the channel spacing difference at the wavelength extremes in this case is
small (about 3%), an equal-wavelength approximation is close to an equal-frequency
spacing definition.
10.1.2 Standards for WDM
Because WDM is essentially frequency division multiplexing at optical carrier
frequencies, the WDM standards developed by the International Telecommunication
Union (ITU) specify channel spacing in terms of frequency [7–10]. A key reason for
selecting a frequency spacing that is fixed, rather than a wavelength spacing that is
constant, is that when locking a laser to a particular operating mode, it is the frequency
of the laser that is fixed. Recommendation G.692 was the first ITU-T specification